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## [$POJ$ $1041$ $John's$ $trip$](http://poj.org/problem?id=1041)
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### 一、题目大意
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多组数据，输入$x，y，z$，表示结点$x$和结点$y$之间有一条序号为$z$的边，如果这个 **无向图** 中存在欧拉回路，就 **输出字典序最小的欧拉回路**，如果不存在欧拉回路就输出 `Round trip does not exist. `。当输入`0 0`表示一组数据输入结束，题目保证了图的连通性。
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给出一张无向图，要求从起点开始遍历一遍所有的边，最后再回到起点，题目要求输出任意一组方案
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#### 细节
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- 起点不是点$1$，而是第一条边中两个端点中较小的一个点
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- 给出的$x$ $y$ $z$代表的是点$x$到点$y$由$id$为$z$的边连接
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- 最后答案要求输出的是边的$id$
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### 二、解题思路
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#### 欧拉回路
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对于一个图可以从一个顶点沿着边走下去，每个边只走一次，所有的边都经过后回到原点的路。
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#### 欧拉回路判定
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<font color='red' size=4><b>
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- 无向图存在欧拉回路 $\Leftrightarrow$ 每个顶点的度是偶数
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- 有向图存在欧拉回路 $\Leftrightarrow$ 每个顶点的出度等于入度（就是出去的边数等于进来的边数）
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</b></font>
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#### 解题步骤
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先根据欧拉路的定义判断是否存在欧拉路，如果存在的话再 **求字典序最小的欧拉路**，一定是以边$1$为起始的欧拉路，然后将每个结点的边按序号从大到小排序，从而保证$dfs$的时候得到的是字典序最小的欧拉路。
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> **解释**：链式前向星是与$dfs$配合的，栈的原理，由大到小排序，最后走的就是小的，反着打出来就是由小到大
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### 三、$do \ while$版本
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```cpp {.line-numbers}
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#include <cstdio>
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#include <cstdlib>
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#include <cstring>
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#include <algorithm>
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#include <cmath>
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#include <stack>
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#include <queue>
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#include <vector>
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#include <map>
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#include <string>
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#include <iostream>
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using namespace std;
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const int N = 2010, M = N * N;
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// 声明结构体
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struct Node {
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int a, b, c;
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bool const operator<(const Node &t) const {
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return c > t.c;
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}
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} g[N];
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int st[M]; // 某条边是不是访问过，无向图一般用于处理成对就换的边
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int d[N]; // 度
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int res[M], rl; // 路径数组
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int m; // m条边
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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void dfs(int u) {
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for (int i = h[u]; ~i; i = h[u]) { // 删边优化
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h[u] = ne[i];
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if (st[i]) continue;
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st[i] = st[i ^ 1] = 1; // 成对变换
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dfs(e[i]);
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// 记录路径
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res[++rl] = w[i];
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}
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("POJ1041.in", "r", stdin);
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#endif
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int a, b, c;
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while (true) {
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memset(d, 0, sizeof d);
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memset(st, 0, sizeof st);
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memset(h, -1, sizeof h);
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memset(res, 0, sizeof res);
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idx = rl = m = 0;
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// 本题的输入挺奇怪的，需要使用while(true)+do while的方式来读取最方便
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scanf("%d%d", &a, &b);
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if (a == 0 && b == 0) exit(0);
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do {
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scanf("%d", &c);
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g[m].a = a, g[m].b = b, g[m++].c = c;
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scanf("%d%d", &a, &b);
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} while (a && b);
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// 针对边的边号进行由小到大排序
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sort(g, g + m);
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for (int i = 0; i < m; i++) {
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int a = g[i].a, b = g[i].b, c = g[i].c;
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add(a, b, c), add(b, a, c);
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d[a]++, d[b]++;
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}
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int flag = 0;
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for (int i = 0; i < m; i++) {
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int a = g[i].a, b = g[i].b;
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if ((d[a] & 1) || (d[b] & 1)) {
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flag = 1;
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break;
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}
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}
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if (!flag) {
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dfs(1);
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// 逆序输出序列
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for (int i = rl; i; i--) printf("%d ", res[i]);
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} else
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cout << "Round trip does not exist.";
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puts("");
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}
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return 0;
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}
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```
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### 四、$while$版本
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```cpp {.line-numbers}
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#include <cstdio>
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#include <cstdlib>
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#include <cstring>
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#include <algorithm>
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#include <cmath>
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#include <stack>
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#include <queue>
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#include <vector>
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#include <map>
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#include <string>
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#include <iostream>
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using namespace std;
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const int N = 2010, M = N * N;
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// 声明结构体
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struct Node {
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int a, b, c;
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bool const operator<(const Node &t) const {
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return c > t.c;
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}
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} g[N];
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int st[M]; // 某条边是不是访问过，无向图一般用于处理成对就换的边
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int d[N]; // 度
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int res[M], rl; // 路径数组
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int m; // m条边
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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void dfs(int u) {
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for (int i = h[u]; ~i; i = h[u]) { // 删边优化
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h[u] = ne[i];
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if (st[i]) continue;
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st[i] = st[i ^ 1] = 1; // 成对变换
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dfs(e[i]);
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// 记录路径
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res[++rl] = w[i];
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}
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("POJ1041.in", "r", stdin);
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#endif
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int a, b, c;
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while (true) {
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memset(d, 0, sizeof d);
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memset(st, 0, sizeof st);
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memset(h, -1, sizeof h);
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memset(res, 0, sizeof res);
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idx = rl = m = 0;
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scanf("%d%d", &a, &b);
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if (a == 0 && b == 0) exit(0);
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scanf("%d", &c);
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g[m].a = a, g[m].b = b, g[m++].c = c;
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while (true) {
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scanf("%d%d", &a, &b);
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if (a == 0 && b == 0) break;
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scanf("%d", &c);
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g[m].a = a, g[m].b = b, g[m++].c = c;
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}
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// 针对边的边号进行由小到大排序
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sort(g, g + m);
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for (int i = 0; i < m; i++) {
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int a = g[i].a, b = g[i].b, c = g[i].c;
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add(a, b, c), add(b, a, c);
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d[a]++, d[b]++;
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}
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int flag = 0;
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for (int i = 0; i < m; i++) {
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int a = g[i].a, b = g[i].b;
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if ((d[a] & 1) || (d[b] & 1)) {
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flag = 1;
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break;
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}
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}
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if (!flag) {
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dfs(1);
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// 逆序输出序列
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for (int i = rl; i; i--) printf("%d ", res[i]);
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} else
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cout << "Round trip does not exist.";
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puts("");
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}
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return 0;
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}
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```
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